Complex Number1

Definition

Complex Number

Calculation

Applications And Problems

Contents 

Definition of Complex Number is presented by Mr Minh Ngoc Le

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Calculation of Complex Number is presented by Mr Hoai Nam Tran

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Applications of Complex Number is presented by Mr Anh Tuan Nguyen

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Problems of Complex Number is Presented by Ms Hoai Thuong Phan

Contents 

Definition: Where was ‘i’ hiding? Definition of Complex Number Venn Graph for all Numbers Graphical Presentation

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Calculation: Some basic method Complex Conjugate Polar Form Another basic method from Polar Form

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Applications of Complex Number Express cos nα or sin nα by using complex number Express cos^n α or sin^n α by using Complex number

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Problems of Complex Number

The complex numbers To make many of the rules of mathematics apply universally we need to enlarge our number field.

We desire that every integer has an inverse element, we accept the existence of rational numbers. If we desire every polynomial equation to have root(s) equal in number to its highest variable power, we must extend the real number field R to a larger field C of complex numbers.

Where was ‘i’ hiding? You may remember being told that you can't take the square root of a negative number. That's because you had no numbers that, when squared, were negative. Squaring a negative number always gives you a positive. So you couldn't very well square-root a negative and expect to come up with anything sensible. Now, however, you can take the square root of a negative number, but it involves using a new number to do it. At one time, nobody believed that any "real world" use would be found for this new number, other than easing the computations involved in solving certain equations, so the new number was viewed as being a ‘pretend number’ invented for convenience sake.

A complex number A complex number is an ordered pair of real numbers (a,b). We call ‘a’ the real part and ‘b’ the imaginary part of the complex number. We write that new number as a + bi. The '+' is used to indicate the sign of the imaginary number part. The real number part represented by ‘a’ which can be either positive or negative.

Examples : 2 - 4i -3 + 5i -5 + 3/4i These are examples of numbers that we say are strictly complex.

VENN DIAGRAM Representation

 Since

all number belong to the Complex number field, C, all number can be classified as complex. The Real number field, R, and the imaginary numbers, i, are subsets of this field as illustrated below. Complex Numbers a + bi Real Numbers Imaginary Numbers a + 0i 0 + bi

Graphical representation of a complex number A

complex number has a representation in a plane. Simply take the x-axis as the real numbers and an y-axis as the imaginary numbers. Thus, giving the complex number a + bi the representation as point P with coordinates (a,b).

Graphing a Complex Number  Therefore,

complex numbers can be represented by a two dimensional graph.  Here we see the graph of the complex number 3 – 2i.

1.Addition and Multiplied method

Example

Subtracted and divided method

Example

Complex Conjugate

Example

Polar Formula

Another form of basic method

Applications to trigonometric indentities

example

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“An example is always better than thousands explanations” Albert Einstein

example

Prove that: The two square roots of a+bi are (x +yi) and -(x +yi) with y = sqrt((r - a)/2) and x = b/(2.y)        

Solution We are looking for all real numbers x and y so that (x + iy)(x + iy) = a + ib <=> x² - y² + 2xyi = a + bi <=> x² - y² = a and 2xy = b Because b is not 0, y is not 0 and so <=> x²- y²= a and x = b/(2y) <=> b²/(4y²) - y² = a and x = b/(2y)

(1) (2) (3)

(4)

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The first equation of (4) gives us y and the second gives the corresponding x-value. Let t = y² in the first equation of (4) then 4t² + 4at - b²= 0 (5) Let r = modulus of a + bi The discriminant = 16(a ²+ b²) = 16r² We note the roots as t1 and t2. <=> t1 = (- a + r)/2 and t2 = (- a - r)/2 (6) Since y is real and r > a, t1 > 0 and gives us values of y. Since the product of the roots of (5) is (-b²/4) < 0 , t2 is strictly negative.

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So we find two values of y. We note these values y1 and y2. y1 = sqrt((r - a)/2) and y2 = -sqrt((r - a)/2) (7) The corresponding x values are x1 = b/(2.y1) and x2 = b/(2.y2) (8) Note that the two solutions are opposite complex numbers.

solve z² = -4i 

Solution Let x+iy a square root of -4i.

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The modulus of -i is r = 4

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y = sqrt((r - a)/2) = sqrt(2) x = b/(2.y) = (-4)/(2.sqrt(2)) = -sqrt(2)

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The two solutions are -sqrt(2)+isqrt(2) +sqrt(2)-isqrt(2)

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Solve ax²+bx+c=0 with a,b,c€R, a<>0 and b²-4ac < 0.     

Solution Discriminant= b²-4ac=i²(4ac-b²) The roots are: x1= (-b+isqrt(4ac-b²))/(2a) x2= (-b-isqrt(4ac-b²))/(2a)

Find the polar representation of (i-sqrt(3))        

Solution The modulus of (i-sqrt(3)) is 2. (i-sqrt(3)) = 2.( -sqrt(3)/2 + (1/2)i ) Say, the argument is alpha. cos(alpha) = -sqrt(3)/2 sin(alpha) = (1/2) Choose alpha = 5 pi/6 (i-sqrt(3)) = 2.(cos(5.pi/6) + i sin(5.pi/6))

Calculate ( cos(2)+ i sin(2) + 1)ⁿ       

Solution Appealing on trigonometric formulas we have (1 + cos(2)) = 2 cos² (1) sin(2)=2.sin(1).cos(1) (1 + cos(2)+ isin(2)) = 2 cos² (1) + i.2.sin(1).cos(1) = 2.cos(1) (cos(1) + i sin(1)) (1 + cos(2)+ isin(2))ⁿ = 2ⁿ .cosⁿ (1) .(cos(n) + i sin(n))

Given: n is a positive integer. z is a complex number with modulus 1, such that z²ⁿ is not -1. Show that zⁿ/(1+z²ⁿ) is a real number        

Solution Since z is a complex number with modulus 1, we can write z = (cos(t) + i sin(t)) zⁿ= (cos(n t) + i sin(n t)) 1 + z²ⁿ = 1 + cos(2n t) + i sin(2n t) = 2 cos²(nt) + 2 i sin(n t) cos(n t) = 2 cos(nt). (cos(nt) + i sin(nt)) zⁿ/(1+z²ⁿ)=1/(2 cos(nt)) and this is real.

Presented by:Hoai Nam Tran (Alex) Minh Ngoc Le (Daniel) Anh Tuan Nguyen (Tom) Hoai Thuong Phan (Alice)

Coppy right by Alex from wikipedia.com